I have a domain for an integral given by: $$\sqrt{x^2+y^2} \le z \le \sqrt{2-x^2-y^2}$$
I've sketched the domain on a plane and can see the shape it makes (a sort of diamond shape) but in my question I'm asked to find the the two surfaces forming the boundary of this domain and I'm not entirely sure what this means/what this is. I can't really picture any boundaries.
I'll put an image of my plane and if anyone could explain/show what these boundaries are it would really help, thanks in advance.
You already sketched it: since both inequalities must be observed, your domain is the intersection of the inside of the cone with the inside of the hemisphere. So the surface forming the boundary of this domain is dependent on $z$, since for one given $z$, only one condition can be active with equality (boundary) whereas for the other, strict inequality holds (inside). So the two surfaces are (1) the cone: $\sqrt{x^2+y^2} =z $ for $0 \le z \le 1$ and (2) the cap $z = \sqrt{2-x^2-y^2} $ for $1 \le z \le \sqrt{2}$.